   # You mix 125 mL of 0.250 M CsOH with 50.0 mL of 0.625 M HF in a coffee-cup calorimeter, and the temperature of both solutions rises from 21.50 °C before mixing to 24.40 ° C after the reaction. CsOH(aq) + HF(aq) → CsF(aq) + H 2 O( ℓ ) What is the enthalpy of reaction per mole of CsOH? Assume the densities of the solutions are all 1.00 g/mL, and the specific heat capacities of the solutions are 4.2 J|/g · K. ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 5, Problem 38PS
Textbook Problem
333 views

## You mix 125 mL of 0.250 M CsOH with 50.0 mL of 0.625 M HF in a coffee-cup calorimeter, and the temperature of both solutions rises from 21.50 °C before mixing to 24.40 °C after the reaction.CsOH(aq) + HF(aq) → CsF(aq) + H2O(ℓ)What is the enthalpy of reaction per mole of CsOH? Assume the densities of the solutions are all 1.00 g/mL, and the specific heat capacities of the solutions are 4.2 J|/g · K.

Interpretation Introduction

Interpretation:

The enthalpy change for the reaction per mole of CsOH has to be determined.

Concept Introduction:

Standard enthalpy of the reaction,ΔrHo, is the change in enthalpy that happens when matter is transformed by a given chemical reaction, when all reactants and products are in their standard states.

Enthalpy of the reaction,ΔrH, is the change in enthalpy that happens when matter is transformed by a given chemical reaction

Heat energy required to raise the temperature of 1g of substance by 1K..Energy gained or lost can be calculated using the below equation.

q=C×m×ΔT

Where, q= energy gained or lost for a given mass of substance (m), C =specific heat capacity, ΔT= change in temperature.

### Explanation of Solution

Given,

Specific heat capacity of the solution=4.20JK/g

Density of HF=1g/mL

Mass of HF=1g/mL× 50mL =50g

Mass of CsOH=1g/mL × 125mL =125g

Determine the amount of CsOH

50g(1mol÷149.912 g) = 0.333mol

Assume qr+qsol=0

qsol Can be calculated from q=C×m×ΔT, as

qsol=(50g of HF + 125g of CsOH) × 4

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